Plane Equations & Normals - Multivariable Calculus

The angle between the planes $x+y=1$ and $y+z=2$ is $\theta$ . What is the value of $\cos(\theta)$ ?
Answer: $\frac{1}{2}$ When you encounter a question about the angle between two planes, you need to find the angle between their normal vectors. The angle between planes is defined as the acute angle between their normal vectors. To find normal vectors, look at the coefficients of x, y, and z in each plane equation. For the plane $x + y = 1$ , we can rewrite this as $x + y + 0z = 1$ , giving us normal vector $\vec{n_1} = \langle 1, 1, 0 \rangle$ . For the plane $y + z = 2$ , we can write $0x + y + z = 2$ , giving us normal vector $\vec{n_2} = \langle 0, 1, 1 \rangle$ . The cosine of the angle between these vectors uses the dot product formula: $\cos(\theta) = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|}$ Calculate each component: - $\vec{n_1} \cdot \vec{n_2} = (1)(0) + (1)(1) + (0)(1) = 1$ - $|\vec{n_1}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ - $|\vec{n_2}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$ Therefore: $\cos(\theta) = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$ Choice (A) $0$ would mean perpendicular planes, but our dot product isn't zero. Choice (B) $\frac{\sqrt{3}}{2}$ corresponds to a 30° angle, which doesn't match our calculation. Choice (C) $\frac{\sqrt{2}}{2}$ corresponds to a 45° angle and likely comes from incorrectly using just the magnitude $\sqrt{2}$ without proper division. Remember: always extract normal vectors from the coefficients, then apply the dot product formula systematically. The angle between planes always uses their normal vectors, not direction vectors.
The plane $x + 4y + 8z = 16$ intersects the positive coordinate axes at points $A$ , $B$ , and $C$ . What is the area of the triangle $ABC$ ?
Answer: $36$ First, find the points where the plane intersects the coordinate axes. The x-intercept is found by setting $y=0$ and $z=0$ , which gives $x=16$ . So, $A=(16,0,0)$ . The y-intercept is found by setting $x=0$ and $z=0$ , giving $4y=16$ , so $y=4$ . Thus, $B=(0,4,0)$ . The z-intercept is found by setting $x=0$ and $y=0$ , giving $8z=16$ , so $z=2$ . Thus, $C=(0,0,2)$ . Now, we can form two vectors on the sides of the triangle, for example, $\vec{AB} = \langle 0-16, 4-0, 0-0 \rangle = \langle -16, 4, 0 \rangle$ and $\vec{AC} = \langle 0-16, 0-0, 2-0 \rangle = \langle -16, 0, 2 \rangle$ . The area of the triangle $ABC$ is half the magnitude of the cross product of these two vectors. $\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -16 & 4 & 0 \\ -16 & 0 & 2 \end{vmatrix} = \mathbf{i}(8-0) - \mathbf{j}(-32-0) + \mathbf{k}(0 - (-64)) = \langle 8, 32, 64 \rangle.$ The magnitude is $||\langle 8, 32, 64 \rangle|| = \sqrt{8^2 + 32^2 + 64^2} = \sqrt{64 + 1024 + 4096} = \sqrt{5184} = 72$ . The area of the triangle is $\frac{1}{2} ||\vec{AB} \times \vec{AC}|| = \frac{1}{2}(72) = 36$ .
A plane $\mathcal{P}$ contains all points in space that are equidistant from the points $A(-2, 1, 5)$ and $B(4, 3, -1)$ . What is the distance from the origin $(0,0,0)$ to the plane $\mathcal{P}$ ?
Answer: $\frac{\sqrt{19}}{19}$ The plane $\mathcal{P}$ is the perpendicular bisector of the segment $AB$ . The normal vector to the plane, $\vec{n}$ , is the vector from $A$ to $B$ : $\vec{n} = \vec{AB} = \langle 4 - (-2), 3 - 1, -1 - 5 \rangle = \langle 6, 2, -6 \rangle$ . We can use a simpler, parallel vector $\vec{n'} = \langle 3, 1, -3 \rangle$ . The plane passes through the midpoint $M$ of the segment $AB$ : $M = \left( \frac{-2+4}{2}, \frac{1+3}{2}, \frac{5-1}{2} \right) = (1, 2, 2)$ . The equation of the plane is $3(x - 1) + 1(y - 2) - 3(z - 2) = 0$ , which simplifies to $3x - 3 + y - 2 - 3z + 6 = 0$ , or $3x + y - 3z + 1 = 0$ . The distance from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D = 0$ is given by the formula $\frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ . The distance from the origin $(0,0,0)$ to the plane $\mathcal{P}$ is $d = \frac{|3(0) + 1(0) - 3(0) + 1|}{\sqrt{3^2 + 1^2 + (-3)^2}} = \frac{|1|}{\sqrt{9 + 1 + 9}} = \frac{1}{\sqrt{19}}$ . Rationalizing the denominator gives $\frac{\sqrt{19}}{19}$ .
A plane $\mathcal{P}$ passes through the point $(1, 2, -3)$ and is perpendicular to the line of intersection of the planes $x - y + 2z = 4$ and $2x + y - z = 1$ . Which of the following is an equation for the plane $\mathcal{P}$ ?
Answer: $x - 5y - 3z = 0$ Let the two given planes be $\mathcal{P}_1$ and $\mathcal{P}_2$ . Their normal vectors are $\vec{n}_1 = \langle 1, -1, 2 \rangle$ and $\vec{n}_2 = \langle 2, 1, -1 \rangle$ . The direction vector, $\vec{d}$ , of the line of intersection is perpendicular to both normal vectors. We can find $\vec{d}$ by computing the cross product: $\vec{d} = \vec{n}_1 \times \vec{n}_2 = \langle (-1)(-1) - (2)(1), (2)(2) - (1)(-1), (1)(1) - (-1)(2) \rangle = \langle -1, 5, 3 \rangle$ . The desired plane $\mathcal{P}$ is perpendicular to this line, so its normal vector $\vec{n}$ is parallel to $\vec{d}$ . We can choose $\vec{n} = \langle -1, 5, 3 \rangle$ . The plane passes through the point $(1, 2, -3)$ , so its equation is given by $-1(x - 1) + 5(y - 2) + 3(z - (-3)) = 0$ . This simplifies to $-x + 1 + 5y - 10 + 3z + 9 = 0$ , which is $-x + 5y + 3z = 0$ . Multiplying by -1 gives the equivalent equation $x - 5y - 3z = 0$ .
A tetrahedron has vertices at $A(1, 0, 0)$ , $B(0, 2, 0)$ , $C(0, 0, 3)$ , and $D(1, 1, 1)$ . What is the equation of the plane containing face $ABC$ , and what is the distance from vertex $D$ to this plane?
Answer: $6x + 3y + 2z = 6$ ; distance = $\frac{5}{7}$ For plane through $A(1,0,0)$ , $B(0,2,0)$ , $C(0,0,3)$ , use vectors $\vec{AB} = (-1, 2, 0)$ and $\vec{AC} = (-1, 0, 3)$ . The normal vector is $\vec{AB} \times \vec{AC} = (-1, 2, 0) \times (-1, 0, 3) = (6, 3, 2)$ . Using point $A(1, 0, 0)$ : $6(x-1) + 3(y-0) + 2(z-0) = 0$ gives $6x + 3y + 2z = 6$ . The distance from $D(1, 1, 1)$ to this plane is $\frac{|6(1) + 3(1) + 2(1) - 6|}{\sqrt{6^2 + 3^2 + 2^2}} = \frac{|6 + 3 + 2 - 6|}{\sqrt{36 + 9 + 4}} = \frac{5}{\sqrt{49}} = \frac{5}{7}$ .
Which statement best describes the intersection of the three planes defined by the equations $x + 2y - z = 3$ , $2x - y + 3z = 1$ , and $8x + y + 7z = 9$ ?
Answer: The planes intersect in a single line. First, we examine the normal vectors of the planes: $\vec{n}_1 = \langle 1, 2, -1 \rangle$ , $\vec{n}_2 = \langle 2, -1, 3 \rangle$ , and $\vec{n}_3 = \langle 8, 1, 7 \rangle$ . No two vectors are scalar multiples, so no two planes are parallel. To determine if the planes intersect at a point, a line, or not at all (forming a prism), we check if the normal vectors are coplanar by calculating their scalar triple product: $\det(\vec{n}_1, \vec{n}_2, \vec{n}_3) = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 8 & 1 & 7 \end{vmatrix} = 1(-7-3) - 2(14-24) - 1(2-(-8)) = -10 + 20 - 10 = 0.$ Since the scalar triple product is zero, the normal vectors are coplanar. This means the planes either intersect in a single line or form a prism. To distinguish these cases, we check for consistency in the system of linear equations. Notice that $\vec{n}_3 = 2\vec{n}_1 + 3\vec{n}_2$ . We check if the same linear combination applies to the constant terms of the plane equations written as $Ax+By+Cz-D=0$ . Let the equations be $E_1$ , $E_2$ , $E_3$ . We check if $2(x+2y-z-3) + 3(2x-y+3z-1) = 0$ corresponds to the third plane. This simplifies to $8x+y+7z-9=0$ , which is exactly the third plane's equation. Since the third equation is a linear combination of the first two, the system is dependent and consistent. Therefore, the three planes intersect in a single line.
Two points on the line $\vec{r}(t) = \langle 1+t, 2-t, 2t \rangle$ are at a distance of $\sqrt{6}$ from the plane $x+2y+z=1$ . What is the distance between these two points?
Answer: $12\sqrt{6}$ A general point on the line is $P(t) = (1+t, 2-t, 2t)$ . The distance from this point to the plane $x+2y+z-1=0$ is given by the formula $D = \frac{|Ax_0+By_0+Cz_0+D_0|}{\sqrt{A^2+B^2+C^2}}$ . $D = \frac{|(1+t) + 2(2-t) + (2t) - 1|}{\sqrt{1^2+2^2+1^2}} = \frac{|1+t+4-2t+2t-1|}{\sqrt{6}} = \frac{|t+4|}{\sqrt{6}}$ We are given that this distance is $\sqrt{6}$ . So, $\frac{|t+4|}{\sqrt{6}} = \sqrt{6}$ , which implies $|t+4|=6$ . This gives two values for $t$ : $t+4 = 6 \implies t_1 = 2$ and $t+4 = -6 \implies t_2 = -10$ . The two points on the line are $P_1 = \vec{r}(2) = \langle 3, 0, 4 \rangle$ and $P_2 = \vec{r}(-10) = \langle -9, 12, -20 \rangle$ . The distance between these two points is the magnitude of the vector connecting them. This vector is $\vec{P_2P_1} = \langle 3 - (-9), 0 - 12, 4 - (-20) \rangle = \langle 12, -12, 24 \rangle$ . The distance is $||\langle 12, -12, 24 \rangle|| = \sqrt{12^2 + (-12)^2 + 24^2} = \sqrt{144+144+576} = \sqrt{864}$ . Simplifying the radical, $864 = 144 \times 6$ , so the distance is $\sqrt{144 \times 6} = 12\sqrt{6}$ . Alternatively, once $t_1$ and $t_2$ are found, the distance is $|t_1 - t_2| ||\vec{d}||$ , where $\vec{d}=\langle 1, -1, 2 \rangle$ is the direction vector of the line. $||\vec{d}||=\sqrt{1^2+(-1)^2+2^2}=\sqrt{6}$ . The distance is $|2 - (-10)|\sqrt{6} = 12\sqrt{6}$ .
Find the equation of the plane that contains the line $x = 1 + 2t, y = -1 + 3t, z = 4t$ and the point $P(2, 1, 1)$ .
Answer: $5x - 2y - z = 7$ First, identify the direction vector of the line, $\vec{v} = \langle 2, 3, 4 \rangle$ . Next, find a point on the line by setting $t=0$ , which gives $P_0(1, -1, 0)$ . To find the normal vector of the plane, we need two non-parallel vectors that lie in the plane. One is $\vec{v}$ , and another can be the vector connecting the point on the line $P_0$ to the given point $P$ , which is $\vec{w} = \vec{P_0P} = \langle 2-1, 1-(-1), 1-0 \rangle = \langle 1, 2, 1 \rangle$ . The normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{v}$ and $\vec{w}$ , so we can find it using the cross product: $\vec{n} = \vec{v} \times \vec{w} = \langle (3)(1) - (4)(2), (4)(1) - (2)(1), (2)(2) - (3)(1) \rangle = \langle -5, 2, 1 \rangle$ . Using the point-normal form with point $P(2, 1, 1)$ and normal $\vec{n}$ , the equation of the plane is $-5(x-2) + 2(y-1) + 1(z-1) = 0$ , which simplifies to $-5x + 10 + 2y - 2 + z - 1 = 0$ , or $-5x + 2y + z = -7$ . Multiplying by -1 gives $5x - 2y - z = 7$ .
Three planes have equations $x + 2y - z = 4$ , $2x - y + 3z = 1$ , and $3x + y + 2z = k$ . For what value of $k$ do the three planes intersect in exactly one point, and what happens when $k = 7$ ?
Answer: Any $k \neq 5$ ; when $k = 7$ , the planes have no common intersection Set up the augmented matrix and row reduce. The coefficient matrix has determinant $\begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & 2 \end{vmatrix} = 1(-2-3) - 2(4-9) - 1(2+3) = -5 + 10 - 5 = 0$ . Since the coefficient matrix is singular, the system either has no solution or infinitely many solutions. Row reducing the augmented matrix shows that the system is consistent (has solutions) when $k = 5$ , giving infinitely many solutions (intersection in a line). For $k \neq 5$ , the system is inconsistent (no solutions). When $k = 7$ , the system is inconsistent, so the planes have no common intersection point.
The line $L$ is parameterized by $\vec{r}(t) = \langle t, 2-t, 1+t \rangle$ . The plane $\mathcal{P}$ contains the line $L$ and is orthogonal to the plane $x+y+z=1$ . Which of the following is an equation for plane $\mathcal{P}$ ?
Answer: $x - z = -1$ Let the desired plane be $\mathcal{P}$ and the given plane be $\mathcal{P}_1$ . Let $\vec{n}$ be the normal vector to $\mathcal{P}$ and $\vec{n}_1$ be the normal vector to $\mathcal{P}_1$ . From the equation of $\mathcal{P}_1$ , we have $\vec{n}_1 = \langle 1, 1, 1 \rangle$ . The direction vector of the line $L$ is $\vec{d} = \langle 1, -1, 1 \rangle$ . Since the plane $\mathcal{P}$ contains the line $L$ , its normal vector $\vec{n}$ must be orthogonal to the direction vector $\vec{d}$ . Since plane $\mathcal{P}$ is orthogonal to plane $\mathcal{P}_1$ , their normal vectors are orthogonal. Therefore, $\vec{n}$ must be orthogonal to both $\vec{d}$ and $\vec{n}_1$ . We can find $\vec{n}$ by computing the cross product: $\vec{n} = \vec{d} \times \vec{n}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}(-1-1) - \mathbf{j}(1-1) + \mathbf{k}(1-(-1)) = \langle -2, 0, 2 \rangle.$ We can use a simpler normal vector parallel to this, such as $\vec{n'} = \langle -1, 0, 1 \rangle$ . To find the equation of the plane, we need a point on it. Since it contains the line $L$ , we can use any point from the line. For $t=0$ , we get the point $P_0(0, 2, 1)$ . Using the point-normal form: $-1(x-0) + 0(y-2) + 1(z-1) = 0$ , which simplifies to $-x + z - 1 = 0$ , or $x - z = -1$ .
Two planes are tangent to the sphere $x^2 + (y-1)^2 + (z+2)^2 = 4$ and are parallel to the plane $2x - 2y + z = 5$ . What is the distance between these two tangent planes?
Answer: $4$ The equation of the sphere is $x^2 + (y-1)^2 + (z+2)^2 = 4$ . Its center is $C(0, 1, -2)$ and its radius is $r = \sqrt{4} = 2$ . The two tangent planes are parallel to the plane $2x - 2y + z = 5$ , so they share the same normal vector $\vec{n} = \langle 2, -2, 1 \rangle$ . The two tangent planes will be located on opposite sides of the sphere's center, each at a distance equal to the radius. The total distance between these two parallel planes is therefore the diameter of the sphere, which is $2r = 2(2) = 4$ . Alternatively, one can find the equations of the planes. They are of the form $2x - 2y + z = D$ . The distance from the center $C(0, 1, -2)$ to such a plane must be the radius, 2. Using the distance formula: $\frac{|2(0) - 2(1) + 1(-2) - D|}{\sqrt{2^2 + (-2)^2 + 1^2}} = 2$ $\frac{|-4 - D|}{\sqrt{9}} = 2 \implies |D+4| = 6$ This gives two values for $D$ : $D+4 = 6 \implies D=2$ and $D+4 = -6 \implies D=-10$ . The two planes are $2x - 2y + z = 2$ and $2x - 2y + z = -10$ . The distance between two parallel planes $Ax+By+Cz=D_1$ and $Ax+By+Cz=D_2$ is $\frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}}$ . So the distance is $\frac{|2 - (-10)|}{\sqrt{2^2+(-2)^2+1^2}} = \frac{12}{3} = 4$ .
What are the coordinates of the reflection of the point $Q(3, 1, 0)$ across the plane $x - y + z = 5$ ?
Answer: $(5, -1, 2)$ Let $Q=(3,1,0)$ . The normal vector to the plane is $\vec{n} = \langle 1, -1, 1 \rangle$ . Let's find a point $P_0$ on the plane, for instance, by setting $y=0, z=0$ , we get $x=5$ , so $P_0=(5,0,0)$ . Let $\vec{v} = \vec{P_0Q} = \langle 3-5, 1-0, 0-0 \rangle = \langle -2, 1, 0 \rangle$ . The vector from the reflection point $Q_{refl}$ to $Q$ is twice the projection of the vector $\vec{QP_0}$ onto $\vec{n}$ , or $-2\text{proj}_{\vec{n}}\vec{v}$ . The projection vector is given by: $\text{proj}_{\vec{n}}\vec{v} = \frac{\vec{v} \cdot \vec{n}}{||\vec{n}||^2} \vec{n}$ The dot product is $\vec{v} \cdot \vec{n} = (-2)(1) + (1)(-1) + (0)(1) = -3$ . The squared magnitude is $||\vec{n}||^2 = 1^2 + (-1)^2 + 1^2 = 3$ . So, $\text{proj}_{\vec{n}}\vec{v} = \frac{-3}{3} \langle 1, -1, 1 \rangle = \langle -1, 1, -1 \rangle$ . The position vector of the reflection is $\vec{OQ}_{refl} = \vec{OQ} - 2 \text{proj}_{\vec{n}}\vec{v}$ . So, $\vec{OQ}_{refl} = \langle 3, 1, 0 \rangle - 2\langle -1, 1, -1 \rangle = \langle 3, 1, 0 \rangle - \langle -2, 2, -2 \rangle = \langle 5, -1, 2 \rangle$ . The reflected point is $(5, -1, 2)$ .

Start this quiz